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ECE 4211 UCONN-ECELW3 Lecture Week 3-2 (02042016)

Chapter 2 Notes P-n and n-p junction Review: Forward and Reverse biasing

Energy Band DiagramsAvalanche and Zener Breakdown

Circuit models; Junction and Diffusion CapacitanceHeterojunctions

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Carrier recombination and photon emission: (see also LEDs and Lasers)•Radiative (photon emission) and non-radiative transitions: Since hole belongs to the valence band and the electrons to the conduction band, the recombination of an electron-hole pair results in energy release. When the energy is released as a photon, the transition is called radiative. On the other hand, when the energy released as phonons (or lattice vibrations) it is nonradiative, and causes heating of the material. • •Number of photons produced per second (in a region defined by x) depends on the current Ip(x). The total number of photons produced per second in n-region is Ip(xn)* hq/q. Quantum efficiency hq defines the ratio of probability of a radiative transition. It is expressed as hq = (1/tr)/[(1/tr) + (1/tnr)], here, tr and tnr are the radiative and non-radiative lifetimes.

The quantum efficiency is higher in the direct energy gap semiconductors than in indirect energy gap semiconductors. It takes longer for an electron-hole pair to recombine in indirect energy gap semiconductors as they need the assistance of a phonon to conserve the momentum.

k vector

Energy

E-K diagram of an indirectsemiconductor

Energy Gap Egk vector

Energy

E-K diagram of an directsemiconductor

Energy Gap Eg

•Both energy and momentum are conserved in a downward transition, such as that occurs when an electron-hole pair recombines. In direct energy gap semiconductors, the recombination via photon emission is more likely than in indirect semiconductors, such as Si and Germanium. In indirect energy gap semiconductors, the electrons and holes have different momentum or wave vectors (k). [Momentum is related to k via the Planck’s constant; p=h/l=(h/2p)k].

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Extraction of emitted photons

•Photons have comparable energy (E=hn) ) to electrons and holes. However, they have very small momentum when compared to them. As a result, photon absorption and emission, involving electrons and holes, are shown by vertical lines on the E-k diagram (or energy band diagram). By contrast, lattice vibrations or phonons travel with the speed of sound, and have very small energy (~20 to 80 milli electron volt). But their momentum is comparable to that of electrons and holes. As a result phonons play a crucial role in transitions involving photons in indirect transitions.

•Extraction of generated photons: once the photons are generated they travel in different directions. Depending on the application, sometime we want to stay inside a cavity (lasers) or move along waveguides (photonic integrated circuits) and other times we want to extract them out (light-emitting diodes, LEDs). See the section on wave guiding layers to configure LEDs as lasers.

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2.5.7 Energy band diagrams under forward and reverse biasing:Fig. 10 shows energy band diagrams for equilibrium, forward bias and reverse biasing conditions. We outline steps to obtain these diagrams:Step 1. Draw a straight line and call it Fermi energy under equilibruui (dashed line).Step 2. Draw two vertical lines representing junction. Note that the equilibrium width Wo (left figure) is larger than forward bias diode (center Wf) and smaller than the reverse biased diode (Wr right). Step 3. Draw new Fermi level in the neutral p-semiconductor with respect to the n-semiconductor. There is no change in equilibrium or left figure. Under forward bias, p-region is qV f lower than the equilibrium value as shown in the center figure. In forward bias of 0.2V, p-side is positive, but the electron energy band diagram is 0.2eV below the equilibrium Fermi level. The case is opposite in the reverse bias of -0.2V. Here, p-side is Vr = -0.2V negative with respect to n-side, but the new Fermi level is 0.2eV higher than the equilibrium level. Step 4. Draw valence band and conduction band on the p-side using new Fermi level.Step 5. Join conduction bands and valence bands between the two vertical lines.

Band bendingqVbi=0.754 eV

p

1.1eV=Eg

Ec

W0=0.9928 µm

Ef

n

Forward bias Vf=0.2V

Wf=0.576µm

qVf

Band bandingq(Vbi+Vr)=0.954VeV

Ec

Ef

Band bending Vbi-Vf=0.554 eV

Ef (new)

Wr=1.11µm

Ef (new)

Reverse Bias Vr=-0.2V

Ev

Ev

Ev

Ec

Ev Ev

Ec Ec

Ev

p n

p n

qVR

Equilibrium

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2.5.4Alternate approach to derive I-V equation using stored charge:In this approach, the charge stored in the p- and n-regions is computed. The current is stored charge divided by the average lifetime of the injected minority electrons in the p-region of the n-p diode. In an n-p diode,

In(x=xp) = Qn/tn (82)

here, Qn is the excess minority charge stored in the p-region. Current In(xp) is obtained by integrating qdn(x) [to find charge Qn] and diving by life time tn.

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2.5.5 Diffusion capacitance:

When the diode is forward biased, the stored minority carrier charges in the neutral regions varies with the applied voltage, and contribute to a capacitive component known as diffusion capacitance Cdiff.

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Junction width and Junction Capacitance p.106

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2/1))((2

DA

DAbior

DA

DA

NqNNNV

NNANqN

=Q

Substituting Eq. 40A for Wo, we get

And under forward bias

CDiffCJ

GD

10

Reverse bias junction

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Carrier distribution in a reverse biased p-n junction

pn(x)=dp(x) + pno= pno(1- )e L)x--(x

p

n

for x>xnr

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2

pLp

xp dd

x>x> for eBeA=p(x) nL)x-(x

L)x--(x

p

n

p

nd

The second boundary condition shows that B=0 otherwise we have a solution which is not realistic.

The first boundary condition givesA = -pno

This gives

x>x> for epno=p(x) nL)x--(x

p

nd

Here, xn should be replaced by xnr

dxxdp

D-q=(x)Jn

pp)( Jp(x) = q Dp pno (-1/Lp) e L

)x--(x

p

n

e L)x--(x

p

n

= - [(q Dp pno)/Lp]

The current density at x= xn or xnr Jp(xn) = - [(q Dp pno)/Lp],

Jp(x) = Jp(xn) e L)x--(x

p

n , for x>xn

Jp(xn) = - [(q Dp pno)/Lp],

Jn(x) = -[(q Dn npo)/Ln] e n

p

L)x++(x

Thus the current under reverse biasing is

J = - Js, where Js is the reverse saturation current Js.

UConn-ECE4211, F. JainWeek 3-Lecture 2 02102015

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2.6 Avalanche and Zener Breakdown in Reverse-Biased p-n Junctions

Figure 11(a). Reverse biased p+-n junction.

Figure 11. (b) E-field distribution, 11 (c) Electron-hole pair generation by accelerating holes

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Avalanche Breakdown p-n Junctions

Figure 11. (c) Current flow in diodes; (top) p-n junction with width W and p-i-n with W’ (bottom).

nno

npo pno

W’

(ii) p-i-n diode with a wider junction width W’ > W.

E = Electric field

xE

ppo e

h

Holes and electrons accelerate in the electric field (E) of the depletion region. When kinetic energy (KE) > 3/2Eg, an electron-hole pair (EHP) is created. So one electron or hole causes generation of an EHP. This is shown here.

v

i

W

PIp(0)

+In(0)N

+-

--

+

0

(i) p-n diode

e

Avalanche Breakdown

qJ+

qJ+

dxJd

q1-=0 pnn

J+J=dxJd

pnn

)J+J-(=dxJd

pnp

The avalanche generation rate I+I pn

(107)

α = electron ionization coefficient and β = hole ionization coefficient.

•Case I: β << α and β=0 dx=IId x

0n

n(x)I(0)I

n

n

e(0)I=e(0)I=(x)I yn

dxnn

x0

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Avalanche Breakdown p-n Junctions

Fig. 11(d) . Variation of the electron hole current components in the depletion region.

(107)

•Case I: β << α and β=0

e(0)I=e(0)I=(x)I yn

dxnn

x0

I

Ip(0)

In(0)

0 W XIp(W)

In(W)Itotal

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2.6.2 Zener Breakdown from energy band perspective

(107)

Fig. 12 shows schematically the energy band diagram under equilibrium and reverse baised conditions. Wr represents the width of the potential barrier for electrons on p-side. Electron can tunnel through the potential barrier Wr if it is around 50-100Å. This field-assisted tunneling gives rise to Zener breakdown. If the field assisted tunneling is negligible, then the device follows avalanche breakdown.

Ec

Ef p

p-Sin+

Ec

EV

Eg=1.1eV

e

Ev

qE=Felectron

Direction of Electric

n+

Efnl

(Fermi level in the conduction band due to n+ doping)

n* side is 10 Volts positive or on energy band diagram, it is q10 V = -10eV negative.

Fig. B Energy band diagram of a n+-P diode.

New electron-hole pair formed due to impact ionization of a hot electron (electron gaining energy in the depletion region) and giving part of its energy in creating electron-hole pair.

WRqVR= -10eV

Equilibrium Fermi level on n side

n+ p

E(x) in WR

Fig. A E field in n+ p diode

-xnR -xpR

E field

x

VBR=EcWR/2

If

-10VoltVBR

Fig. C Increased reverse current in the breakdown region of characteristic.

VIs

VR=-10V

h

0

NNq)N+N()VV2(

=WrDA

DAorrbi21

1/2

DA

DA0rrbizr NqN

)N(Nε)εV2(VW

2/1

D

0r

qNεε)(2

rbi

zrVV

WNA>> ND

I

Is

VZener

Fig. 15 I-V of a Zener diode.

V

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2.7 Tunnel diodes:Figure 17 shows the energy band diagram of a tunnel diode. Since diode n-side (left) and p-side (right side) are heavily doped, Fermi levels are above Ec and below Ev. That is inside the bands. They are not in the energy gap as we usually show. The thin barrier also increases the tunneling probability. Fig. 18A shows tunneling in more detail.

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2.7 Tunnel diodes: page 132

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Application of Tunnel Junctions: Solar Cells

GaAs Substrate Zn-doped

GaInP 0.07um p=3*10^17 cm-3 [Zn]

GaAs 3.5um p=8*10^16 cm-3 [Zn]

GaAs 0.1um n=1*10^18 cm-3 [Se]GaInP 0.1um n=1*10^18 cm-3 [Se]

GaAs 0.01um n=1*10^19 cm-3 [Se]GaAs 0.01um p=8*10^19 cm-3 [C]

GaInP 0.6um p=1.5*10^17 cm-3 [Zn]

GaInP 0.1um n=2*10^18 cm-3 [Se]

AlInP 0.025um n=4*10^17 cm-3 [Si]

GaAs 0.5um n=6*10^18cm-3 [se]

Au 3um AR Coat (2 layer)

Front Grids

Contacting layers

(Eg=1.86ev)

Back Contact

Top Cell

Tunnel Cell

Bottom Cell

GaInP/GaAs Tandem Cell

Fig. 18C Tunnel junction interfacing two n-p cells. Here, the bottom n-p GaAs cell and top n-p GaInP cell.

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Application of Tunnel Junctions: FETs

In MOS-FETs, source to drain tunneling takes place when channel lengths are very small and barrier heights are small.

VDS

P - SiN +

nn

N +

Ec

Ec'Ef’

Ev

N-Si

30 A Fig. 20C Electrons tunneling from the inversion channel to the gate in a MOSFET.

Example Fowler-Nordheim tunneling (The potential barrier has a slope due to the presence of the voltage drop across oxide SiO2).

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Heterojunctions: Single heterojunciton

1.85eV

1.43eVEc = 0.252 eV

Ev = 0.168 eV

p-AlGaAs n-GaAs

Ec

Ev

W

0-Xp0 Xn0

Ec

Ev

Ef

Eg ~ 1.9 eV

Eg = 1.424 eV

Fig. 36. Energy band diagram for a p-n heterojunction.

Fig. 35. Energy band diagram per above calculations. N-p heterojunction.

2.9.5. Single heterojunctions: Energy band diagrams for N-AlGaAs – p-GaAs and P-AlGaAs/n-GaAs heterojunctions under equilibrium

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Energy band diagram: Double Heterojunction

N AlGaAs p GaAs

0

-xn xp

x

VF=1.0 Volt

d

P-Al

GaAs

Fig. 39. A forward biased NAlGaAs-pGaAs-PAlGaAs double heterojunction diode.

Eg2(AlGaAs)

Eg2(p-AlGaAs)=1.85eV

Ev(n-p)

Ev(pGaAs-pAlGaAs)

p-P heterojunction1.85eV=

Ec(pGaAs-pAlGaAs)

Eg1(GaAs)

Ec(n-p)

Ec

Ev

0-Xn0 Xp0

N-p heterojunction

Fig. 42 Energy band diagram of a NAlGaAs-pGaAs-PAlGaAs double heterostructure diode.

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Built-in Voltage in Heterojunctions

Fig. 33. Energy band diagram line up before equilibrium.

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Built-in Voltage in Heterojunctions

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Built-in Voltage in Heterojunctions Cont.

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2.9.3.2. Built-in Voltage Method II: Gauss' Law

or

poA

or

NoD

pompNmNbi

xNxNq

xExE21=V

1

2

2

2

0

21

21

(145)

2.9.4. Forward-Biased NAlGaAs-pGaAs Heterojunction

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0

N-AlGaAs pGaAs

X

pe

pno

ne

npo

Concentrations

np(x)=n(x)dn(x)

xp-xn

Fig.34 Carrier concentrations in an n-p heterojunction

x>x> for nL1=

xdnd

pn

22

2

ddn-(x)n=n popd

(73)

eB+eA=n-(x)n=n(x) L

xx

L

xx-

pop n

p

n

p )()( d

x>x> for e1)-e(n=en=n(x) pL

)x--(xkTVq

poL

)x--(x

n

pf

n

p

d (161)

dxnd

qD+=(x)J nnd

x>x> for e1)-e(L

EnDq=(x)J pL

)x-(x-kTVq

n

gponn n

pf

)( 1

Hole diffusion from pGaAs to the N-AlGaAs,

(164)

Electron diffusion from N-AlGaAs to the p-GaAs side,

->x>x- for e1)-e(p=ep=p(x) NL)x++(x

kTVq

NoL)x++(x

P

Nf

P

Nd

(162)

dxpd

D-q=(x)J ppd

e1)-e(L

pDq=(x)J L

)x+(x+

kTVq

p

AlGaAsNOPP P

Nf)(

I-V Equation

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1)-e(L

pDq+

LnDq

=)x(-J+)x(J=J kTVq

P

NOP

n

ponNPpn

f

)1( ,

2)(

,

2)(

kT

qVf

AlGaAsDP

AlGaAsiP

GaAsAn

GaAsin eNL

nD+NL

nDq=J

Next we substitute the values of npo and pNo in Eq. 107

1)-e(J=J kTVq

sf

,

2)(

,

2)(

AlGaAsDP

AlGaAsiP

GaAsAn

GaAsinS NL

nD+NL

nDq=J (170)

)1( 23

22

11

,

2)(

,

2)(

kTqVfKT

E

pn

pn

AlGaAsDP

GaAsiP

GaAsAn

GaAsin eemmmm

NL

nD+NL

nDq=Jg )1(

,

2)(

kTqVf

GaAsAn

GaAsinn e

NL

nDq=J

I-V Equation and Current Density Plot

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)1( ,

2)(

kTqVf

GaAsAn

GaAsinn e

NL

nDq=J

)1( 23

22

11

,

2)(

kTqVfKT

E

pn

pn

AlGaAsDP

GaAsiPP ee

mmmm

NL

nDq=Jg

)1( 122

3

22

11

,

2)(

,

2)(

kTqVfKT

EE

pn

pn

AlGaAsDP

GaAsiP

GaAsAn

GaAsin eemmmm

NL

nD+NL

nDq=Jgg

)(23

11

222

),(

2),(

12

1

2 kT

EE

pn

PN

TEi

TEigg

g

g emmmm

n

n

Here, we have used the energy gap difference Eg=Eg2-Eg1. From Eq. 174 we can see that the second term, representing hole current density Jp which is injected from p-GaAs side into N-AlGaAs, and it is quite small as it has [exp-(Eg/kT)] term. As a result, J ~ Jn(xp), and it is

In(xp)/e=0.011mA

Ln

x

Ip(-xn)/e

-xn 0

In(xp)=0.051mA

Lp=1.414x10-4cm 10µm

Ip(-xn)=5.97pA

xpNot to Scale

Ip(x<-xn)

I=Ip(-xn)+In(xp)

I-Ip(-xn)Ip(x>xp)

Fig. 38B Current density plots.